A) \[{{I}_{1}}={{I}_{2}}=\frac{Blv}{6R},I=\frac{Blv}{3R}\]
B) \[{{I}_{1}}=-{{I}_{2}}=\frac{Blv}{R},I=\frac{2Blv}{R}\]
C) \[{{I}_{1}}={{I}_{2}}=\frac{Blv}{3R},I=\frac{2Blv}{3R}\]
D) \[{{I}_{1}}={{I}_{2}}=I=\frac{Blv}{R}\]
Correct Answer: C
Solution :
[c]: Emf induced across PQ is \[\varepsilon =Blv\]. |
The equivalent circuit diagram is as shown in the figure. |
Applying Kirchhoff?s first law at junction Q, we get |
\[I={{I}_{1}}+{{I}_{2}}\] ...(i) |
Applying Kirchhoff?s second law for the closed loop PLMQP, we get |
\[-{{I}_{1}}R-IR+\varepsilon =0\] |
\[{{I}_{1}}R+IR=Blv\] ...(ii) |
Again, applying Kirchhoff?s second law for the closed loop PONQP, we get |
\[-{{I}_{2}}R-IR+\varepsilon =0\] |
\[{{I}_{2}}R+IR=Blv\] ...(iii) |
Adding equations (ii) and (iii), we get |
\[2IR+{{I}_{1}}R+{{I}_{2}}R=2Blv\] |
\[2IR+R({{I}_{1}}+{{I}_{2}})=2Blv\] |
\[2IR+IR=2Blv\] (Using (i)) |
\[3IR=2Blv\] |
\[I=\frac{2Blv}{3R}\] ...(iv) |
Substituting this value of I in equation (ii), we get |
\[{{I}_{1}}=\frac{Blv}{3R}\] |
Substituting the value of I in equation (iii), we get |
\[{{I}_{2}}=\frac{Blv}{3R}\] |
Hence, \[{{I}_{1}}={{I}_{2}}=\frac{Blv}{3R},I=\frac{2Blv}{3R}\] |
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